A potential difference of 220 V is applied across a rheostat of 12000 Ω . The voltmeter V has a resistance of 6000 Ω and distance BC is one-fourth of the distance from A to B . The error in the reading of voltmeter is approximately -

Question

A potential difference of 220 V is applied across a rheostat of 12000 Ω . The voltmeter V has a resistance of 6000 Ω and distance BC is one-fourth of the distance from A to B . The error in the reading of voltmeter is approximately - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rp9vbzecoggu.png" /> (A) 27 % (B) 10 % (C) 40 % (D) 15 %

Tardigrade Admin · Accepted Answer

It total voltage across AB is 220 then voltage across CB should be (220/4)= 55 V (voltage to be measured) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-rp9vbzecjvs8.png" /> Vâ² =(220 × 2000/11000)= 40 V (voltage being measured) ∴ % error=(V - V â²/V)× 100=(55 - 40/55)× 100=27 %

Q. A potential difference of $220 V$ is applied across a rheostat of $12000 Ω$ . The voltmeter $V$ has a resistance of $6000 Ω$ and distance $BC$ is one-fourth of the distance from $A$ to $B$ . The error in the reading of voltmeter is approximately -

$27%$

$10%$

$40%$

$15%$

It total voltage across $A B$ is $220$ then voltage across $CB$ should be $\frac{220}{4} = 55 V$ (voltage to be measured)

$V' = \frac{220 \times 2000}{11000} = 40 V$ (voltage being measured)
$∴ % error = \frac{V - V '}{V} \times 100 = \frac{55 - 40}{55} \times 100 = 27%$

Q. A potential difference of 220V is applied across a rheostat of 12000Ω . The voltmeter V has a resistance of 6000Ω and distance BC is one-fourth of the distance from A to B . The error in the reading of voltmeter is approximately -

27%

10%

40%

15%